Expression for Angular Frequency

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Discussion Overview

The discussion revolves around the derivation of the expression for angular frequency in an L.C.R. circuit, specifically the formula w = (1/LC - R²/4L²)¹/². Participants explore the underlying differential equations and methods for solving them, as well as the relationships between various circuit components.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about the derivation of the expression for angular frequency in an L.C.R. circuit.
  • Another suggests that solving the differential equation based on Kirchhoff's loop rule could lead to the expression.
  • A participant proposes that the expression resembles a determinant and questions whether a differential equation for current can be derived from it.
  • Some participants present alternative forms of the angular frequency, indicating different approaches to solving the problem.
  • There is a discussion about the nature of the differential equation derived from the circuit components, with references to standard methods for solving such equations.
  • Clarifications are made regarding the notation used in differential equations, particularly the distinction between first and second derivatives.
  • One participant expresses uncertainty about the application of certain methods to the equations being discussed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the derivation of the angular frequency expression, with multiple competing views and methods presented throughout the discussion.

Contextual Notes

Some participants express uncertainty regarding the steps involved in solving the differential equations, and there are unresolved aspects related to the specific forms of the equations and their solutions.

zorro
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How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R2/4L2)1/2

(I came across this while solving a problem)
 
Last edited:
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You solve the differential equation for kirchhoffs loop rule.
 
Hi Abdul! :smile:
Abdul Quadeer said:
How does this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R2/4L2)1/2

(I came across this while solving a problem)

Well, without knowing anything about circuits, we can guess that it probably has something to do with the fact that that obviously looks like a determinant,

of Lx2 + Rx + 1/C = 0 …​

can you get a differential equation that looks like that, maybe for current? :wink:
 
I got the same equation for w in place of x. On solving,

w = (-R +/- (R2 -4L/C)1/2)/2L

I did not get any differential equation.
 
Abdul Quadeer said:
I got the same equation for w in place of x. On solving,

w = (-R +/- (R2 -4L/C)1/2)/2L

Are you sure? that's not the same ω as in your original post …
Abdul Quadeer said:
How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R2/4L2)1/2

I suspect that the R/2L is a separate exponential part

(my differential equation in I was from combining CVC = Q (so CdVC/dt = I), VR = IR, VL = LdI/dt, and d/dt(VC + VR + VL) = 0)
 
tiny-tim said:
(my differential equation in I was from combining CVC = Q (so CdVC/dt = I), VR = IR, VL = LdI/dt, and d/dt(VC + VR + VL) = 0)

That will give a second order differential equation, right?
Your expression does not contain any w.
 
It gives I as a function of t …

part of that solution will be harmonic, and ω will be the frequency of the harmonic part. :smile:
 
I got
I/C + RdI/dt + Ld2I/dt2 = 0

How to solve this?
 
Hi Abdul! :smile:

(just got up :zzz: …)
Abdul Quadeer said:
I got
I/C + RdI/dt + Ld2I/dt2 = 0

How to solve this?

Don't you know the standard method for these equations? …

write D = d/dt, so the equation becomes

(LD2 + RD + 1/C)I = 0,

which you can factor to

(D - R/2L + i√(1/LC + R2/4L2))(D - R/2L - i√(1/LC + R2/4L2))I = 0 …

carry on from there :smile:
 
  • #10
Its d2I/dt2 not (dI/dt)2 :wink:
 
  • #11
That's ok, that is what D2 means …

D2(I) = D(D(I)) = d/dt(d/dt(I)) = d/dt(dI/dt) = d2I/dt2 :smile:
 
  • #12


I have solved questions in D.E. like -
1 + (dy/dx)2 = xdy/dx

by taking dy/dx = p

Do we use p2 for both (dy/dx)2 and d2y/dx2 ?
 
  • #13
no …

that p equation has an xdy/dx,

the D is for equations with just dy/dx :smile:
 

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